太原市2024年高三年级模拟考试（一）物理参考答案及评分建议二、选择题:本题共8小题，每小题6分，共48分。在每小题给出的四个选项中，第14～18题只有一项符合题目要求，第19～21题有多项符合题目要求。全部选对的得6分，选对但不全的得3分，有选错的得0分。题号1415161718192021选项ADBBCBDABAC三、非选择题：共62分。22.（6分）（）（分）（分）10.400222�11222��2�1（2）游标卡尺测量遮光条宽度−时存在误差、滑块与轨道间存在摩擦、空气阻力影响、测量A、B点到水平桌面的高度时存在误差等（答案合理均可得分）（2分）23.（11分）（1）3.371~3.374（1分）（2）×1（1分）欧姆（1分）9.0（1分）I(Rr)（3）C（1分）E（1分）202（1分）9.1（1分）9×10-4（2分）I1I2（4）等于（1分）24.（12分）解：（1）升温后有6cm长的水银柱进入左侧竖直细管，水银柱高6cm，空气柱长21cm，76cmHg，封闭气体的压强为ℎ2=�2=·······�··0··=········································�··2··························（2分）�2+7�0�cℎm2H=g··�·0·················································································（2分）�（23=）升温前水银柱高10cm，258K，空气柱长10cm，76cmHg，封闭气体的压强为ℎ1=�1=�1=�0=····�··1·········································································（2分）�0+8�6�cℎm1H=g·�··1·················································································（2分）1�=·····················································································（2分）�1�1��2�2��1=�2441K·······················································································（2分）�2=-1-25．（15分）解：(1)弹簧与质量为7m的小球组成的系统机械能守恒····················································································（2分）7∆��=2���弹簧由原长压缩为原长的一半，弹性势能增加································（1分）72（2）小球P与弹簧组成的系统机械能守恒���=m········································································（1分）712∆��=2���2�1P�1与=Q7发�生�弹性碰撞,以右为正，碰后速度分别为，''········································�·1·····�·2···························（1分）''112��=��+��····································································（1分）121'21'22��1=2��1+2��2=0'�1=························································································（1分）'�（23）7轻�绳�断开时与竖直方向的夹角为θ，小球的速度为3�（分）mgcosθ=2·················································································1�3Q从最低点�摆2�动到轻绳断开处mg2l(1+cosθ)=·························································（1分）121‘2−2��3−2��2cosθ=，θ=60012轻绳断开后，Q将做斜抛运动竖直方向，以上为正�3=��···················································································（1分）3y3�2l=(1+�cosisnθ�)=tgt2·······························································（1分）13水−平方向，以�左s为in正�−2···················································································（1分）�=3x=t�3·c··o·s··�····················································································（1分）��3xl··························································································（1分）�轻=绳在3断开时水平长度为lQ落到B点时，Q与P之间2�s的in距�离=Δ30··············································（1分）�=-2-26.（18分）（1）由右手定则可知导体棒在B1中电流方向由P到Q························································（2分）导体棒在B2中电流方向由Q到P························································（2分）（2）PQ水平跃入轨道的速度为v1=cosθ······················································································（1分）1P�Q水�平跃入导轨后，切割磁感线，C2处于充电状态，PQ稳定后做匀速直线运动，速度为v2···················································································（1分）222Q�=C=2�·�·�·······················································································（1分）根据动�量2定理，以右为正=mv2-mv1········································································（2分）Q−=�2��·∆··�·····················································································（1分）Q=�∆�···············································································（1分）�2�2��